# The sign of an interesting sum involving a Dirichlet character

Let $$\chi_{q}$$ be a primitive Dirichlet character with modulus $$q$$ (see definition at wikipedia ). For example for $$q=5$$ we have

\begin{aligned} \chi_{5,1}&=(1, 1, 1, 1, 0),\\ \chi_{5,2}&=(1, i, -i, -1, 0),\qquad\qquad \text{(1)}\\ \chi_{5,3}&=(1, -1, -1, 1, 0),\\ \chi_{5,4}&=(1, -i, i, -1, 0),\\ \end{aligned}

We construct a sum involving Dirichlet character $$\chi_{q}$$ as

\begin{aligned} Q(\chi_{q})&=\sum_{k=1}^{q-1}k\chi_q(k)\qquad\qquad \text{(2)} \end{aligned}

Proposition A: For a complex $$\chi_q$$, like $$\chi_{5,2},\chi_{5,4}$$, \begin{aligned} &\color{red}{\text{sign}(\mathrm{Re}\chi_{q}(-1))}Q(\mathrm{Re}\chi_{q})>0,\qquad\text{(3)}\\ &\color{red}{\text{sign}(\mathrm{Im}\chi_{q}(-2))}Q(\mathrm{Im}\chi_{q})>0.\qquad\text{(4)}\\ \end{aligned}

We are seeking a proof or a reference of the proof for this proposition.

This problem is related to a partial answer of another problem titled sign unchanged for Dirichlet polynomials? that we posted earlier at math.stackexchange.com. The motivation of studying this problem is also mentioned there.

• Thanks!. I changed the title as you suggested!
– mike
Aug 7 '19 at 6:00
• Sure. Yes. $Q(\mathrm{Re}\chi_q)=\mathrm{Re}(Q(\chi_q))$.
– mike
Aug 7 '19 at 6:18
• Assertions (3) and (4) are both false. There is a character modulo 7 for which $Q(\chi_q) = 0$. Even if you weaken the assertion a non-strict inequality $\ge 0$ rather than $> 0$, you can find characters of modulus $\le 20$ for which (3) and (4) are negative. Did you do any checking at all before you posted this conjecture here? Aug 7 '19 at 8:31
• @DavidLoeffler: You beat me by 58 seconds! Aug 7 '19 at 8:33
• @DavidLoeffler: Thanks a lot for the comment. I am sorry that I have not done enough testing.
– mike
Aug 7 '19 at 16:06

(3) and (4) are false in general, even if we weaken $$>$$ to $$\geq$$. Let $$\zeta:=e^{i\pi/8}$$ be a primitive $$16$$-th root of unity, and let $$\chi$$ be the unique primitive Dirichlet character modulo $$17$$ satisfying $$\chi(3)=\zeta^5$$. Then $$\chi(-1)=-1$$, and $$(\chi(1),\chi(2),\chi(3),\chi(4),\chi(5),\chi(6),\chi(7),\chi(8))=(1,\zeta^6,\zeta^5,\zeta^{12},\zeta^9,\zeta^{11},\zeta^7,\zeta^2).$$ However, \begin{align*}Q(\chi)&=\sum_{k=1}^{16} k\chi(k)=\sum_{k=1}^{8} (2k-17)\chi(k)\\[6pt] &=-(15+13\zeta^6+11\zeta^5+9\zeta^{12}+7\zeta^9+5\zeta^{11}+3\zeta^7+\zeta^2)\\[10pt] &\approx \ 8.84701161719 - 4.91203840222\,i. \end{align*} So $$Q(\chi)$$ has positive real part, even though $$\chi(-1)=-1$$. This contradicts (3). Similarly, if we change $$\zeta^5$$ to $$\zeta^3$$ in the definition of $$\chi$$, we get a counterexample to (4).
• Thanks a lot for the detailed answer! I noticed that you used $\chi(k)=-\chi(17-k)$. What is this (anti)symmetry? Best
• @mike: $\chi(17-k)=\chi(-k)=\chi(-1)\chi(k)=-\chi(k)$. Aug 7 '19 at 16:10